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2 years ago

If 25 ml of 1.50M LiOH are neutralized by 15ml of nitric acid what is the molarity of HNO3

2 answers:

8 0

LiOH + HNO3 --> H2O + LiNO3
OH- + H+ --> H2O
OH - = 1.5M * .025L = .0375 mol
.0375mol/.04L = 0.9375M
Since it's at the equivalence point, OH- = H+ concentration

[HNO3] = 0.9375M

nasty-shy [4]2 years ago

8 0

Answer:

Ca = 2.5M

Explanation:

Volume of Acid (Va) = 15ml

Concentration of acid (Ca) = ?

Volume of base (Vb) = 25

Concentration of base (Cb) = 1.50M

The balanced chemical equation is given as;

LiOH + HNO3 --> LiNO3 + H2O

The formular relaing the parameters is given as;

(CaVa) / (CbVb) = Na/Nb

Where Na = Number of moles of Acid = 1

Nb = Number of moles of Base = 1

Making Ca subject of interest;

Ca = (Na * Cb * Vb) / (Nb * Va)

Ca = (1 * 1.50 * 25) / (1 * 15)

Ca = 37.5 / 15

Ca = 2.5M

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3 years ago

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How many grams of NO can be produced if 204 g of NO2 is mixed with 58.1 g of H2O?

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Answer:

44.4 grams of NO can be produced

Explanation:

Step 1: Data given

Mass of NO2 = 204 grams

Molar mass NO2 = 46.0 g/mol

Mass of H2O = 58.1 grams

Molar mass H2O = 18.02 g/mol

Step 2: The balanced equation

3NO2 + H2O→ 2HNO3 + NO

Step 3: Calculate moles NO2

Moles NO2 = 204 grams / 46.0 g/mol

Moles NO2 = 4.43 moles

Step 4: Calculate moles H2O

Moles H2O = 58.1 grams / 18.02 g/mol

Moles H2O = 3.22 moles

Step 5: Calculate limiting reactant

For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO

NO2 is the limiting reactant. It will completely be consumed (4.43 moles). H2O is in excess. there will react 4.43 /3 = 1.48 moles. There will remain 3.22 - 1.48 = 1.74 moles

Step 6: Calculate moles NO

For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO

For 4.43 moles NO2 we'll have 4.43/3 = 1.48 moles NO

Step 7: Calculate mass NO

Mass NO = 1.48 moles * 30.01 g/mol

Mass NO = 44.4 grams

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2 years ago

If a 7.0 mL sample of vinegar was titrated to the stoichiometric equivalence point with 7.5 mL of 1.5M NaOH, what is the mass pe

NNADVOKAT [17]

First we need to calculate the number of moles of NaOH titrated.

molar concentration = number of moles / solution volume (liter)

number of moles = molar concentration × solution volume

number of moles of NaOH = 1.5 × 0.0075 = 0.01125 moles

Then we look at the chemical reaction:

CH $_{3}$ -COOH + NaOH = CH $_{3}$ COONa + H $_{2}$ O

We can see that 1 mole of acetic acid is reacting with one mole of sodium hydroxide. Then we can conclude that 0.01125 moles of sodium hydroxide reacts with 0.01125 moles of acetic acid.

Now we can get the mass of acetic acid:

number of moles = mass (grams) / molecular mass (g/mol)

mass = number of moles × molecular mass

mass of acetic acid = 0,01125 × 60 = 0.675 g

We assume that the density of the vinegar = 1 g/mL, so the mass percent of acetic acid is:

concentration of acetic acid = (mass of acetic acid / mass of vinegar) × 100

concentration of acetic acid = (0.674 / 7) × 100 = 9.6 %

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2 years ago